How does one arrive at this equality?

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SUMMARY

The discussion centers on demonstrating the equality of the Dirac delta function expressed as an integral over momentum states. Specifically, it establishes that the Dirac delta function, δ(q' - q), can be represented as \(\frac{1}{2\pi} \int dp \, e^{i p (q' - q)}\). The relationship between position and momentum eigenstates is clarified through the expression \(\langle q | p \rangle = e^{ipq}\). The factor of \(\frac{1}{2\pi}\) is acknowledged as a crucial component, although its derivation remains unclear to the participants.

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  • Understanding of Dirac delta function properties
  • Familiarity with quantum mechanics concepts, specifically position and momentum eigenstates
  • Knowledge of Fourier transforms and their applications in quantum mechanics
  • Basic grasp of path integral formulation in quantum physics
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  • Research the derivation of the Dirac delta function from Fourier transforms
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Tac-Tics
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I'm reading about the path integral formulation.

How do you show that:

[tex]\delta(q' - q) = \frac{1}{2\pi} \int dp e^{i p (q' - q)}[/tex]

with δ as the Dirac delta, q and q' as two position eigenstates, and (I'm only guessing) p as an iterator over the set of momenta.

I'm not sure the relationship between q and p that makes this work and I'm also not sure where that [tex]\frac{1}{2 \pi}[/tex] factor comes from.
 
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Ah, reading down half a page, they give the relationship:

[tex]\langle q \mid p \rangle = e^{ipq}[/tex]

I believe that's what I was looking for.

(I just wish they'd put this stuff in the right order! It's hard not having a professor to bug.)

(But I'm still unsure where that 1/2pi comes from).
 

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